Problem: You have found the following ages (in years) of 6 seals. Those seals were randomly selected from the 42 seals at your local zoo: $ 6,\enspace 4,\enspace 12,\enspace 1,\enspace 23,\enspace 3$ Based on your sample, what is the average age of the seals? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 42 seals, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{6 + 4 + 12 + 1 + 23 + 3}{{6}} = {8.2\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {4.84} + {17.64} + {14.44} + {51.84} + {219.04} + {27.04}} {{6 - 1}} $ {s^2} = \dfrac{{334.84}}{{5}} = {66.97\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{66.97\text{ years}^2}} = {8.2\text{ years}} $ We can estimate that the average seal at the zoo is 8.2 years old. There is also a standard deviation of 8.2 years.